Proving that $I+A$ is invertable when $A$ is nilpotent: What intuition leads to a particular approach?

In an answer to this question, it has been suggested to consider the following: $$(I+A)(\sum_{j=0}^n(-A)^j)$$

Through a series of algebraic operations, it can be shown that $$\sum_{j=0}^n(-A)^j$$ is in fact the inverse of $$I+A$$.

How would we have known to multiply by $$\sum_{j=0}^n(-A)^j$$? If there isn't an identity or formula that would indicate such a multiplication is a reasonable avenue of inquiry, then how would we otherwise derive $$\sum_{j=0}^n(-A)^j$$?

• Just think that multiplying by some kind of sum of the powers of $\;A\;$ could help since $\;A^n=0\;$ for some natural $\;n\;$ ... For example, if $\;A^2=0\;$ , then $\;(I+A)(I-A)=I-A^2=I\;$ . This and, of course, the fact that $\;I\;$ commutes with any matrix (of the same order) makes things very simple. – DonAntonio Jul 20 at 17:30
• To come up with the proof, it's best to start with a different (more structural) proof: Working in the quotient ring $\mathbb{Z}\left[A\right]/\left(I+A\right)$, we have $I \equiv -A \mod I+A$, so that $I$ is nilpotent modulo $I+A$ (since $A$ is nilpotent, and thus so is $-A$); but this means that $I^k \equiv 0 \mod I+A$ for some $k$, and therefore $I \equiv 0 \mod I+A$, and thus $I+A$ is invertible (since $1 = 0$ only holds in trivial rings). Now, unravel the use of quotient rings and congruences in this proof, extracting an explicit (or recursive) formula for the inverse. – darij grinberg Jul 20 at 20:44

If $$z\in\mathbb C$$ and $$\lvert z\rvert<1$$, then$$\frac1{1+z}=1-z+z^2-z^3+\cdots$$In other words,$$(1+z)\left(1-z+z^2-z^3+\cdots\right)=1.$$This, together with the fact that $$A^n=0$$ if $$n\gg1$$, should make you think that it would be a good idea to try to prove that an inverse of $$\operatorname{Id}+A$$ is $$1-A+A^2-A^3+\cdots$$ (which is a finite sum, in this case).
I guess a natural way of thinking about this stems from the formula for a geometric series for real numbers: if $$|x| < 1$$, then \begin{align} \sum_{j=0}^\infty x^j = \dfrac{1}{1-x} \end{align} Now, replacing $$x$$ with $$-x$$ gives the formula \begin{align} (1+x)^{-1} = \dfrac{1}{1+x} = \sum_{j=0}^{\infty} (-x)^j \end{align} So, it might be natural to try this for matrices as well. And now if $$A$$ is nilpotent, then on the RHS, we only have a finite number of summands (in fact at most $$n$$ summands if $$A$$ is $$n \times n$$). Hence, one might expect that $$\sum_{j=0}^n (-A)^j$$ is the inverse matrix of $$I+A$$. Then, a simple computation verifies that it actually is.
Recall that $$x^{2n+1}+y^{2n+1}=(x+y)(x^{2n}-x^{2n-1}y+x^2y^{2n-2}-..+y^{2n})$$
Now, since $$I$$ and $$A$$ commute, the same formula holds for $$x=I$$ and $$y=A$$. Therefore, $$I+A^{2n+1}=(I+A)(I-A+A^2-A^{3}+...+A^{2n})$$ Now use the fact that $$A^n=0$$ implies $$A^{n}=A^{n+1}=...=A^{2n+1}=0$$